Answer
$\frac{m(a+h)+b-(ma+b)}{h}=\frac{ma+mh+b-ma-b}{h}=\frac{mh}{h}=m$
$f'(a)=\lim \limits_{h \to 0}m$, which equals to $m$.
Work Step by Step
The algebraic derivative of a function can be described as:
$f'(a)=\lim \limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Here, $f(a)=ma+b$
By substituting, we get:
$\frac{m(a+h)+b-(ma+b)}{h}=\frac{ma+mh+b-ma-b}{h}=\frac{mh}{h}=m$
$f'(a)=\lim \limits_{h \to 0}m$, which equals to $m$.
